Initially, the velocity vector is [tex]\langle 38cos(40^{\circ}),38sin(40^{\circ}) \rangle=\langle 29.110, 24.426 \rangle[/tex]. At the same height, the x-value of the vector will be the same, and the y-value will be opposite (assuming no air resistance). Assuming perfect reflection off the ground, the velocity vector is the same. After 0.2 seconds at 9.8 seconds, the y-value has decreased by [tex]4.9(0.2)^2[/tex], so the velocity is [tex]\langle 29.110, 24.426-0.196 \rangle = \langle 29.110, 24.23 \rangle[/tex].
Converting back to direction and magnitude, we get [tex]\langle r,\theta \rangle=\langle \sqrt{29.11^2+24.23^2},tan^{-1}(\frac{29.11}{24.23}) \rangle = \langle 37.87,50.2^{\circ}\rangle[/tex]
